$h(x)=-(x-5)(x-13)$ 1) What are the zeros of the function? Write the smaller $x$ first, and the larger $x$ second. $\text{smaller }x=$
Explanation: $\begin{aligned} -(x-5)(x-13)&=0 \\\\ x-5=0&\text{ or }x-13=0 \\\\ x={5}&\text{ or }x={13} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $x$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }x\text{-coordinate}&=\dfrac{({5})+({13})}{2} \\\\ &={9} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $h({9})$ : $\begin{aligned} h({9})&=-({9}-5)({9}-13) \\\\ &=-(4)(-4) \\\\ &=16 \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }x&=5 \\\\ \text{larger }x&=13 \end{aligned}$ The vertex of the parabola is at $(9,16)$